MAT187 L19 - Partial Fractions

#MAT187 PCE
Before we begin, some definitions provided by the PCE:

Rational Functions

Rational functions are defined as $f (x) = \frac{p (x)}{q (x)}$ , where both $p (x)$ and $q (x)$ are polynomials.

Proper Rational Functions

For a rational function $f (x) = \frac{p (x)}{q (x)}$ , if the degree of $p (x)$ (numerator) is less than that of $q (x)$ (denominator), then it is known as a proper rational function.

For example:

Function	Rational?	Proper rational?	Justification
$f (x) = \frac{e^{x}}{x^{2} + 5 x + 3}$	❌	❌	Numerator is not a polynomial.
$g (x) = \frac{5 x^{2} - 10 x + 3}{7 x^{4} + 5 x - 1}$	✅	✅	Both numerator and denominator are polynomials, and degree of numerator is less than that of the denominator.
$h (x) = \frac{9 x^{4} - 13 x - 20}{3 x^{4} - 3 x + 4}$	✅	❌	Both numerator and denominator are polynomials, but their degrees are the same.

Partial Fraction Decomposition (PFD)

Partial fraction decomposition is the process of splitting a proper rational function into two separate rational functions that are then easier to integrate.

Example

\int \frac{3 x}{x^{2} - x - 2} d x = \int (\frac{1}{x + 1} + \frac{2}{x - 2}) d x = \ln | x + 1 | + 2 \ln | x - 2 | + C

Note that PFD only works on proper rational functions.

Steps for PFD

Break apart the denominator into two separate simpler functions. These are known as Distinct Linear Factors because the $1$ and $- 2$ are distinct from each other.
1. $\frac{3 x}{x^{2} - x - 2} = \frac{3 x}{(x + 1) (x - 2)}$
Then, split the result into the sum of two separate rational functions, with the numerator expressed by $A$ and $B$ .
1. $\frac{3 x}{(x + 1) (x - 2)} = \frac{A}{x + 1} + \frac{B}{x - 2}$
Then, cross multiply (if that's the right term; idk just look at the math, you tell me wtf I'm doing)
1. $\frac{A}{x + 1} + \frac{B}{x - 2} = \frac{A (x - 2) + B (x + 1)}{(x + 1) (x - 2)}$
You should notice that, since the denominator now matches the original function, the numerator must also match. Thus:
1. $A (x - 2) + B (x + 1) = (A + B) x + (- 2 A + B) = 3 x$
Then, we can solve the following linear system to find A and B.
1. $A + B = 3$ , $- 2 A + B = 0$

In some cases, we can also use long division to make PFD a lot easier.

Example

\int \frac{2 x^{2} + 37 x + 144}{x^{2} + 13 x + 40} d x = \int 2 + \frac{11 x + 64}{x^{2} + 13 x + 40} d x

Distinct Quadratic Factors

In some cases when dealing with PFD, you will encounter a rational function whose denominator cannot be further reduced. We call these denominators (the polynomials) irreducible, since we can't reduce the polynomial into two distinct factors.
For PFD to work in this dire situation, we must ensure that the numerator is a polynomial with degree that is 1 less than that of the denominator. For a quadratic denominator, this would look like $\frac{B x + C}{D x^{2} + E x + F}$ .

Example

\begin{aligned} \int \frac{1}{x (x^{2} + 1)} d x & = \int \frac{A}{x} + \frac{B x + C}{x^{2} + 1} d x \\ ⟹ A (x^{2} + 1) + (B x + C) x = 1 \\ ⟹ A + B = 0, A + C = 1 \\ ⟹ A = 1, B = - 1, C = 0 \end{aligned}

Note that $A = B = 0, C = 1$ technically works using our process, but it's also wrong because it gets rid of the $\frac{A}{x}$ term entirely.

\begin{aligned} ∴ \int \frac{1}{x (x^{2} + 1)} d x & = \int \frac{- 1}{x} + \frac{x}{x^{2} + 1} d x \\ = \ln | x | - \frac{1}{2} \ln | x^{2} + 1 | + C \end{aligned}

PFD with Complex Numbers

You can also take irreducible denominators and reduce them with complex numbers. In most cases, you won't want to do this, but in some situations, you will want to. When do you want to do this? I don't really know, this isn't something in the PCEs; it's just something that was briefly mentioned in lecture.

PCE Quiz Questions

Select ALL the terms that we need in the partial fraction decomposition of $\frac{x - 3}{x^{2} - 1}$ .

Option	Correct?	Explanation
$\frac{C}{x^{2} - 1}$	❌	This is technically valid, but not useful since it's the same as the original function given to us.
$\frac{E}{x + 1}$	✅	This is one of the rational functions needed in PFD.
$A x$	❌	This is not a rational function? And we also do not need it as the numerator for distinct quadratic factors because $x^{2} - 1$ is reducible.
$B$	❌	This is just a variable??? Tf? This could literally be anything. If this is correct I'm gonna throw hands with Siefken.
$\frac{D}{x - 1}$	✅	This is one of the rational functions needed in PFD.

Match each rational function on the left to the appropriate PFD setup on the right. You might not need all the options on the right and several lines might have the same answer!

Original Function	Matched Function	Explanation
$\frac{2}{x (x^{2} + 4)}$	$\frac{A}{x} + \frac{B x + C}{x^{2} + 4}$	$x^{2} + 4$ is irreducible so we need to consider [[#Distinct Quadratic Factors]], i.e., $B x + C$ .
$\frac{2}{x (x^{2} - 4)}$	$\frac{A}{x} + \frac{B}{x + 2} + \frac{C}{x - 2}$	$x^{2} + 4$ is reducible so we can break the whole thing into three separate rational functions.
$\frac{x + 2}{x (x^{2} + 4)}$	$\frac{A}{x} + \frac{B x + C}{x^{2} + 4}$	$x^{2} + 4$ is irreducible so we need to consider [[#Distinct Quadratic Factors]], i.e., $B x + C$ .
$\frac{x + 2}{x (x^{2} - 4)}$	$\frac{A}{x} + \frac{B}{x + 2} + \frac{C}{x - 2}$	$x^{2} + 4$ is reducible so we can break the whole thing into three separate rational functions.